Problem: $x^2-3xy+y^3=3$ Find the value of $\dfrac{dy}{dx}$ at the point $(1,2)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{9}{4}$ (Choice B) B $\dfrac{15}{4}$ (Choice C) C $\dfrac{4}{15}$ (Choice D) D $\dfrac{4}{9}$
Solution: We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $x^2-3xy+y^3=3$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} x^2-3xy+y^3&=3 \\\\ \dfrac{d}{dx}(x^2-3xy+y^3)&=\dfrac{d}{dx}(3) \\\\ \dfrac{d}{dx}(x^2)-3\dfrac{d}{dx}(xy)+\dfrac{d}{dx}(y^3)&=0 \\\\ 2x-3\left(1\cdot y+x\cdot\dfrac{dy}{dx}\right)+3y^2\cdot\dfrac{dy}{dx}&=0 \\\\ 2x-3y-3x\cdot\dfrac{dy}{dx}+3y^2\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 2x-3y-3x\cdot\dfrac{dy}{dx}+3y^2\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(3y^2-3x)&=3y-2x \\\\ \dfrac{dy}{dx}&=\dfrac{3y-2x}{3y^2-3x} \end{aligned}$ Now we can plug the point $(1,2)$ into the expression for $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{3y-2x}{3y^2-3x} \\\\ &=\dfrac{3(2)-2(1)}{3(2)^2-3(1)} \gray{x=1,\,\,y=2} \\\\ &=\dfrac{4}{9} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at the point $(1,2)$ is $\dfrac{4}{9}$.